[SOLVED] Designing Coupled Inductor for high gain boost DC-DC Converter

[sudhinpk]

For my mtech thesis I need to design a coupled inductor for my DC-DC boost converter. There are no details provided in paper except they are using coupled inductor: ETD-59, core pc40, Np : Ns =1 : 4 Lm =48 μH; Lk = 0.25 μH. Can anybody help me with the design?

 

[smijesh]

Please provide the complete schematic to understand the requirements.
It is look like a Boost inductor with another (Auxiliary) winding for powering the Driver IC.

Refer

https://www.pca.com/datasheets/EPC3246-XX.pdf

 

[sudhinpk]

Please provide the complete schematic to understand the requirements.
It is look like a Boost inductor with another (Auxiliary) winding for powering the Driver IC.

Refer

https://www.pca.com/datasheets/EPC3246-XX.pdf

I am attaching my paper. Can you please help...

View attachment DC-DC Conveter with High Step up.pdfView attachment DC-DC Conveter with High Step up.pdf

 

[smijesh]

Core EDT59, Material N87 from Epcos
Np = 12Turns,Ns = 1Turn, Air gap = 2mm

 

[sudhinpk]

Core EDT59, Material N87 from Epcos
Np = 12Turns,Ns = 1Turn, Air gap = 2mm

Can you provide me design steps? or any resources which I can refer? I am planning to design this circuit for 100 W.

 

[FvM]

Np = 12Turns,Ns = 1Turn

Doesn't sound like a reasonable step-up converter. Ns has been specified as 4*Np in the original post.

 

[smijesh]

For 100W output
Core EDT59, Material N87 from Epcos
Np = 9Turns,Ns = 36Turn, Air gap = 1mm

 

[sudhinpk]

Can you please specify how you chosen 9 turns as primary?

 

[smijesh]

AL Value of the core = Inductance/ Turns*Turns

Required air gap is 1mm ,Inductance is 48uH.so the AL value of the core will be 550nH

Number of turns in primary = Sq(Inductance /AL value) = 9 Turns

 

[sudhinpk]

Very Very Thanks

 

[sudhinpk]

How can we choose inductance value same as mutual inductance value?

 

[smijesh]

You can calculate the mutual inductance using below equation
Two inductors whose inductances are given as 75mH and 55mH respectively, are positioned next to each other on a common magnetic core so that 75% (K)of the lines of flux from the first coil are cutting the second coil. Calculate the total mutual inductance(M) that exists between them.
M = K *Sq Root(L1*L2)
M = 0.75* Qq Root(75mH*55mH) =48.2mH

 

[sudhinpk]

Thanks. I asked with respect to my paper. They given Lm = 48uH. Is it mutual inductance? Can you help me to calculate values of L1 and L2 for my topology?

 

[smijesh]

Thanks. I asked with respect to my paper. They given Lm = 48uH. Is it mutual inductance? Can you help me to calculate values of L1 and L2 for my topology?

AL Value = 550nH
L primary = 550nH*9*9 = 44.5 uH
L Secondary = 550nH*36*36 = 713uH
Mutual Inductance
M = 0.75* Qq Root(44.5uH*713uH) =133.6uH

 

[sudhinpk]

How do you got AL Value? Is it from the specifications of the core mentioned in the paper?

 

[smijesh]

I got AL value from core datasheet
AL value without air gap = 4170nH Minimum
AL value with 1mm air gap = 550nH 5%

 

[sudhinpk]

Thanks. In my paper Lm=48uH is designed for 400W. Can we select the same for 100 W?

 

[smijesh]

For 400W we consider 2mm air gap but for 100W application we considered only 1mm air gap so it will workout.

 

[sudhinpk]

That's really helpful.Thanks.
For my report I need to show calculation steps for Lm value. When I tried it with volt second balance I got different value (34 mH). Do you have any suggestions for it?